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Need help on easy differential equation

  1. Apr 20, 2005 #1
    I actually haven't had differential equations, but I managed to somehow to register for the class I am in-_-, fortunately it's easy, but I'm still not sure how

    dx/dt = (.8*sqrt(100+x^2)/x

    I figure multiply out the dt, so dx=(.8*sqrt(100+x^2)/x dt, then integrate

    so x=allthatstuff *t, except I'm not sure that's right, should everything be constant like that?
     
    Last edited: Apr 20, 2005
  2. jcsd
  3. Apr 20, 2005 #2
    [tex] \frac{dx}{dt} = \frac{4}{5}\frac{\sqrt{100+x^2}}{5x} [/tex]

    I THINK you subtract the RHS so then its off the form:

    [tex]Mx' + Nx = 0 [/tex]

    I didnt take diff eq either and dont really know much about it, but if you can take it from there, go for it. My next thought would be some kind of integrating factor.
     
  4. Apr 20, 2005 #3

    dextercioby

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    What do you know about separation of variables...?

    Daniel.
     
  5. Apr 20, 2005 #4
    http://pacific.uta.edu/~qiming/Project2.htm


    There's the actual problem, I already did the computational part and all that, now I just need to compare my result with the analytic one, which I don't know how to do:(
     
  6. Apr 20, 2005 #5

    dextercioby

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    It's a simple integration.

    Can u compute

    [tex] \int \frac{\sqrt{x^{2}+100}}{x} \ dx [/tex]...?


    U could use a hyperbolic substitution

    [tex] x=10\sinh t [/tex]

    Daniel.
     
  7. Apr 20, 2005 #6
    err, how did you get that? I wound up with x on top(in which case you can use the sub u=100+x^2

    but I haven't had the class yet so I've really got no clue



    EDIT: My chief concern is no matter what way I try to solve it, I'm not getting a result that really makes sense with the given problem, and certainly doesn't match my numerical answer.
     
    Last edited: Apr 20, 2005
  8. Apr 20, 2005 #7

    James R

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    You have:

    [tex] \frac{dx}{dt} = \frac{4}{5}\frac{\sqrt{100+x^2}}{5x} [/tex]

    Therefore:

    [tex]\int \frac{4}{25}\,dt = \int \frac{x}{\sqrt{100+x^2}} \, dx + c[/tex]

    To do the integral on the right hand side, put [itex]u = 100 + x^2[/itex]. Then

    [tex]du = 2x dx[/tex]

    and the integral becomes

    [tex]\int \frac{1}{2\sqrt{u}} \, du[/tex]

    CAn you do it from there?
     
  9. Apr 20, 2005 #8
    Yah, that's not the problem anymore

    it's 5/4 sqrt(100+x^2) = t, the problem is look at the actual question in that link I posted. The answer doesn't seem to make sense. At least I don't think it does. Well, it doesn't match my numerical answer at least, which means I probably screwed up. *runs off to fix*

    well yah, it doesn't make any sense considering the actual question
     
    Last edited: Apr 21, 2005
  10. Apr 21, 2005 #9

    dextercioby

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    Yes,alright,it's that integral written by James R.Mine would have been a little tricky...

    Daniel.
     
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